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Chapter 3 Design Theory for Relational Databases

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Contents

Functional Dependencies

Decompositions

Normal Forms (BCNF, 3NF)

Multivalued Dependencies (and 4NF)

Reasoning About FD’s + MVD’s

Remember our questions:

Why do we design relations like the example? -- good design

What makes a good relational database schema? -- no redundancy, no Update/delete anomalies,

what we can do if it has flaws? -- decomposition

New Question:

any standards for a good design?

Normal forms: a condition on a relation schema that will eliminate problems

any standards or methods for a decomposition?

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Boyce-Codd Normal Form (BCNF)

We say a relation R is in BCNF if whenever

X ->Y is a nontrivial FD that holds in R, X is

a superkey.

– Remember: nontrivial means Y is not contained

in X.

– Remember, a superkey is any superset of a key

(not necessarily a proper superset).

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Example

Drinkers(name, addr, beersLiked, manf, favBeer)

FD’s: name->addr favBeer, beersLiked->manf

Only key is {name, beersLiked}.

In each FD, the left side is not a superkey.

Drinkers is not in BCNF

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Another Example

Beers(name, manf, manfAddr)

FD’s: name->manf, manf->manfAddr

Only key is {name} .

name->manf does not violate BCNF, but

manf->manfAddr does.

Beers is not in BCNF

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Decomposition into BCNF

Given: relation R with FD’s F.

Aim: decompose R to reach BCNF

Step 1: Look among the given FD’s for a BCNF

violation X ->Y.

– If any FD following from F violates BCNF, then

there will surely be an FD in F itself that violates

BCNF.

Step 2: Compute X +.

– Not all attributes, or else X is a superkey.

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Decomposition into BCNF (cont.)

Step 3: Replace R by relations

with schemas:

1. R1 = X +.

2. R2 = R – (X + – X ).

R-X + X X +-X

R2

R

R1

Decomposition into BCNF (cont.)

Step4: Project given FD’s F onto the two

new relations.

given FD’s F find all implied FD’s pick

up those FD’s which have only attributes

needed.

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Example: BCNF Decomposition

Drinkers(name, addr, beersLiked, manf, favBeer)

F = name->addr, name -> favBeer, beersLiked->manf

Step1: Pick BCNF violation name->addr.

Step2: Closure the left side: {name}+ = {name, addr, favBeer}.

Step3: Decomposed relations:

1. Drinkers1(name, addr, favBeer)

2. Drinkers2(name, beersLiked, manf)

Step4: projecting FD’s to drinker1 and drinker2

Step5: check drinker 1 and drinker2 for BCNF

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Example -- Continued

Given FD’s: name->addr, name -> favBeer, beersLiked->manf

All FD’s: same as given FD’s

For Drinkers1(name, addr, favBeer), relevant FD’s are name->addr and name->favBeer. – Thus, {name} is the only key and Drinkers1 is in BCNF.

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Example -- Continued

For Drinkers2(name, beersLiked, manf), the only FD is beersLiked->manf, and the only key is {name, beersLiked}.

– Violation of BCNF.

beersLiked+ = {beersLiked, manf}, so we decompose Drinkers2 into:

1. Drinkers3(beersLiked, manf)

2. Drinkers4(name, beersLiked)

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Example -- Concluded

The resulting decomposition of Drinkers :

1. Drinkers1(name, addr, favBeer)

2. Drinkers3(beersLiked, manf)

3. Drinkers4(name, beersLiked)

Notice: Drinkers1 tells us about drinkers,

Drinkers3 tells us about beers, and Drinkers4 tells

us the relationship between drinkers and the beers

they like.

Classroom Exercise

Any two-attribute relation R(A,B) is in BCNF

True or False ?

Three Cases:

1) No nontrivial FD’s

2) AB

3) AB, BA

Text book on page 89

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Third Normal Form -- Motivation

There is one structure of FD’s that causes

trouble when we decompose.

R(A,B,C) AB ->C and C ->B.

– Example: A = street address, B = city, C = zip

code.

There are two keys, {A,B } and {A,C }.

C ->B is a BCNF violation, so we must

decompose into R1(AC), R2(BC).

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We Cannot Enforce FD’s

Cannot enforce the FD AB ->C after

decomposition.

Original: R(A,B,C) AB ->C and C ->B.

Decompose into

R1(AC) no FD’s

R2(BC) with C->B

Assume A = street, B = city, and C = zip

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We Cannot Enforce FD’s (cont.)

A C

545 Tech Sq. 02138 545 Tech Sq. 02139

B C

Cambridge 02138 Cambridge 02139

Join tuples with equal C (zip codes).

A B C

545 Tech Sq. Cambridge 02138 545 Tech Sq. Cambridge 02139

FD: A (street) B(city) -> C(zip) is violated by the database as a whole.

CB

keeps

A = street, B = city,

and C = zip

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3NF Let’s Us Avoid This Problem

3rd Normal Form (3NF) modifies the BCNF

condition so we do not have to decompose in

this problem situation.

An attribute is prime if it is a member of any

key.

X ->A violates 3NF if and only if X is not a

superkey, and also A is not prime.

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Example: 3NF

In our problem situation R(A,B,C) with FD’s

AB ->C and C ->B, we have keys AB and

AC.

Thus A, B, and C are each prime.

Although C ->B violates BCNF, it does not

violate 3NF.

R(A,B,C) above is in 3NF, not in BCNF

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BCNF vs. 3NF

conditions example

BCNF

If X ->Y is a nontrivial FD

that holds in R, X is a

superkey.

R(A,B,C)

with AB,

AC

3NF If X ->Y is a nontrivial FD

that holds in R, X is a

superkey, or Y is a prime

R(A,B,C)

with AB -

>C and C

->B.

2 NF: no nonkey attribute is dependent on only a portion of the

primary key. R(A,B,C) with AB, BC

1 NF: every component of every tuple is an atomic value.

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Properties of a decomposition

Elimination of anomalies by a decomposition, it

needs other two properties:

1. Lossless Join : it should be possible to project the

original relations onto the decomposed schema, and

then reconstruct the original.

2. Dependency Preservation : it should be possible to

check in the projected relations whether all the given

FD’s are satisfied.

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Decomposition for 3NF and BCNF

We can get (1: Lossless Join ) with a BCNF

decomposition.

we can’t always get (1) and (2 : Dependency

Preservation ) with a BCNF decomposition.

We can get both (1) and (2) with a 3NF

decomposition.

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Testing for a Lossless Join

If we project R onto R1, R2,…, Rk , can we

recover R by rejoining?

Any tuple in R can be recovered from its

projected fragments.

So the only question is: when we rejoin,

do we ever get back something we didn’t have originally?

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Example

Any tuple in R can be recovered from its

projected fragments.

Not less, not more.

A B C

1 2 3

4 2 5

A B

1 2

4 2

B C

2 3

2 5

A B C

1 2 3

1 2 5

4 2 3

4 2 5

As long as FD bc holds, the

joining of two projected tuples

cannot produce a bogus tuple

⋈

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The Chase Test (an example)

Aim: to prove that a tuple t in the join, using FD’s in F, also to be a tuple in R.

Method: Suppose tuple t comes back in the join.

Then t is the join of projections of some tuples of R, one for each Ri of the decomposition.

Can we use the given FD’s to show that one of these tuples must be t ?

R1(A,B) R2(B,C) R(A,B,C) Decomposed into = Join of R1and R2 Tuple t

(a,b,c)

Belong to

a b c1

a2 b c Because B C, c1= c , therefore a,b,c is in R

textbook on page 98

⋈

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The Chase Test – (method)

1. Start by assuming t = abc… .

2. For each i, there is a tuple si of R that has a,

b, c,… in the attributes of Ri.

3. si can have any values in other attributes.

4. We’ll use the same letter as in t, but with a

subscript, for these components.

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Example: The Chase

Let R = ABCD, the given FD’s be C->D and

B ->A

Suppose the decomposition be AB, BC, and

CD.

Question: Is it a lossless join or not?

R(ABCD) == R1(AB) ⋈ R2(BC) ⋈ R3(CD)

Example: The Chase (cont.)

1. Suppose the tuple t = abcd is the join of

tuples projected onto AB, BC, CD.

2. For each i, there is a tuple s I of R that has

a, b from R1(AB), b,c from R2(BC) and c,d

from R3(CD)

3. a I bI cI dI are any values

Aim: t=abcd is also in the R

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The Tableau

A B C D

a b c1 d1

a2 b c d2

a3 b3 c d d

Use C->D

a

Use B ->A

We’ve proved the second tuple must be t.

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Summary of the Chase Test method

1. If two rows agree in the left side of a FD, make their

right sides agree too.

2. Always replace a subscripted symbol by the

corresponding unsubscripted one, if possible.

3. If we ever get an unsubscripted row, we know any

tuple in the project-join is in the original (the join is

lossless).

4. Otherwise, the final tableau is a counterexample.

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Example: Lossy Join (more tuples)

Same relation R = ABCD and same

decomposition: AB, BC, and CD.

But with only the FD C->D.

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The Tableau

A B C D

a b c1 d1

a2 b c d2

a3 b3 c d

d

Use C->D These three tuples are an example R that shows the join lossy. abcd is not in R. More tuples

These projections rejoin to form abcd.

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Some results

Some decompositions can not keep lossless join

(lossy join).

Use chase method to find out whether the

decomposition is lossy join.

BCNF decomposition is lossless join, sometimes

it can not keep functional dependencies.

Relations with 3NF keep lossless join and also

functional dependencies.

How to decompose relations to reach 3NF?

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3NF Synthesis Algorithm

We can always construct a decomposition into

3NF relations with a lossless join and

dependency preservation.

Need minimal basis for the FD’s:

1. Right sides are single attributes.

2. No FD can be removed.

3. No attribute can be removed from a left side.

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Constructing a Minimal Basis

1. Split right sides.

2. Repeatedly try to remove an FD and see if

the remaining FD’s are equivalent to the

original.

3. Repeatedly try to remove an attribute from

a left side and see if the resulting FD’s are

equivalent to the original.

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3NF Synthesis – method

1. Find a minimal basis for F

2. One relation for each FD in the minimal

basis.

1. Schema is the union of the left and right sides.

2. XA then (XA) is a schema.

3. If no key is contained in an FD, then add

one relation whose schema is some key.

Algorithm 3.26 is on pp.103

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Example: 3NF Synthesis

Relation R = ABCD.

FD’s A->B and A->C. Key is AD

Decomposition: AB and AC from the FD’s,

plus AD for a key.

R is decomposed into R1(AB), R2(AC), R3(AD)

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Another example

Relation R(A,B,C,D,E) and FD’s ABC,

CB,AD

Using 3NF synthesis to decompose:

1) A minimal basis

2) R1(ABC) R2(CB) R3(AD)

3) R2 is a part of R1, delete R2

4) No key is in R1, R3, add a key R4(ABE)

R has two

keys: ABE,

ACE. Add

one of

them

Classroom exercises

Given R(A,B,C,D,E) FD’s ABC, CB, AD

To test R1(ABC), R3(AD), R4(ABE) is in

3NF

To test whether functional dependency

keeps in the R1,R3,R4

To test the decomposition is lossless.

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Why It Works (3NF synthesis)

Preserves dependencies: each FD from a minimal basis is contained in a relation, thus preserved.

Lossless Join: use the chase to show that the row for the relation that contains a key can be made all-unsubscripted variables.

Question:

Why we say “BCNF decomposition” ,”3NF synthesis”?

BCNF decomposition algorithm

Input: relation R + FDs for R

Output: decomposition of R into BCNF relations

w

With “lossless join”

1.Compute keys for R

2.Repeat until all relations are in BCNF:

Pick any R’ with A B that violates BCNF

Decompose R’ into R1(A+) and R2(A, rest)

Compute FDs for R1 and R2

Compute keys for R1 and R2

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3NF synthesis

Input: relation R + FDs for R

Output: decomposition of R into 3NF

relations w

With “lossless join” and keep FD’s

1.Computer key of R

2.Find a minimal basis for F

3.One relation for each FD in the minimal

basis.

4.If no key is contained in an FD, then add one

relation whose schema is some key.

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Summary

Conditions of Norm Forms (BCNF, 3NF)

The way to decompose in order to reach

BCNF

The way to decompose in order to reach 3NF

The way to test the join is lossless join

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